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Q. 10.14.8

Show that the rate of decrease in kinetic energy due to viscosity in a finite volume V of an incompressible fluid is given by

$W=\mu\int_{V}\textbf{w}^2\ dV-\mu\int_{S}\textbf{n}.(\textbf{V}\times\textbf{w})dS$                      (10.14.58)

where S is the boundary of V.
If S is a rigid solid surface at rest, deduce that

$W=\mu\int_{V}\textbf{w}^2\ dV=\int_{v}\Phi dV$                    (10.14.59)

Verified Solution

From (8.6.1), we recall that the kinetic energy of a continuum contained in a volume V is given by

$K=\frac{1}{2}\int_{V}\rho(\textbf{v.v})dV$

Using (8.2.10), we obtain

$\frac{D}{Dt}\int_{V}(\rho\phi)dV=\int_{V}\rho\frac{D\phi}{Dt}dV$             (8.2.10)

$\frac{DK}{Dt}=\frac{1}{2}\int_{V}\rho\frac{D}{Dt}(\textbf{v.v})dV=\int_{V}\rho\textbf{v}.\frac{D\textbf{v}}{Dt}$                    (10.14.60)

For an incompressible viscous fluid, expression (10.14.60) becomes, on using the Navier-Stokes equation (10.14.6),

$\frac{DK}{Dt}=\mu\int_{V}\textbf{v}.\triangledown^2\textbf{v}\ dV-\int_{V}\rho\textbf{v}.\triangledown p\ dV+\int_{V}\rho\textbf{v.b}\ dV$                    (10.14.61)

The last two terms on the righthand side of this expression represents the rate of work done by the pressure and the body force, and the first term represents the contribution of viscosity to the rate of change of kinetic energy. Therefore, if W denotes the rate of decrease in kinetic energy due to viscosity, we have

$W=-\mu\int_{V}\textbf{v}.\triangledown^2\textbf{v}\ dV$                      (10.14.62)

Since the fluid is incompressible, we note by using identities (3.4.27) and (3.4.24) that

$div(\textbf{u}\times\textbf{v})=\textbf{v}.curl\ \textbf{u}-\textbf{u}.curl\ \textbf{v}$           (3.4.24)

$curl\ curl\ \textbf{u}=\triangledown div\ \textbf{u}-\triangledown^2\textbf{u}$              (3.4.27)

$\left.\begin{matrix}\triangledown^2\textbf{v}=-curl\ curl\ \textbf{v}=-curl\ \textbf{w}\\ div(\textbf{v}\times\textbf{w})=\textbf{w}^2-\textbf{v}.curl\ \textbf{w}\end{matrix}\right\}$                    (10.14.63)

In view of these expressions, (10.14.62) becomes

$W=\mu\int_{V}\textbf{w}^2\ dV-\mu\int_{V}div(\textbf{v}\times\textbf{w})dV$                    (10.14.64)

This expression yields (10.14.58), on using the divergence theorem (3.6.1).
If S is a rigid solid surface at rest, then by the no-slip boundary condition we have v = 0 on S. Consequently, the surface integral on the righthand side of (10.14.58) becomes identically 0, and we obtain

$\int_{V}(div\ \textbf{u})dV=\int_{S}(\textbf{u.n})dS$          (3.6.1)

$W=\mu\int_{V}\textbf{w}^2\ dV$                (10.14.65)

which is the first part of (10.14.59). To obtain the other part, we recall expression (10.10.19). For an incompressible fluid, this expression becomes

$\int_{V}\Phi\ dV=\int_{V}\left\{\lambda(div\ \textbf{v})^2+2\mu\frac{D}{Dt}(div\ \textbf{v})^2+\mu\textbf{w}^2\right\}dV\\+2\mu\int_{S}\frac{Dv}{Dt}.\textbf{n}\ dS$            (10.10.19)

$\int_{V}\Phi\ dV=\mu\int_{V}\textbf{w}^2\ dV+2\mu\int_{S}\frac{D\textbf{v}}{Dt}.\textbf{n}\ dS$                  (10.14.66)

If S is a rigid surface at rest, the surface integral in the expression (10.14.66) vanishes and we get

$\int_{V}\Phi\ dV=\mu\int_{V}\textbf{w}^2\ dV$                    (10.14.67)

which is the second part of (10.14.59)